Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-5x+y &= -8 \\ 4x-y &= 6\end{align*}$
Begin by moving the $y$ -term in the second equation to the right side of the equation. $4x = y+6$ Divide both sides by $4$ to isolate $x$ $x = {\dfrac{1}{4}y + \dfrac{3}{2}}$ Substitute this expression for $x$ in the first equation. $-5({\dfrac{1}{4}y + \dfrac{3}{2}}) + y = -8$ $-\dfrac{5}{4}y - \dfrac{15}{2} + y = -8$ Simplify by combining terms, then solve for $y$ $-\dfrac{1}{4}y - \dfrac{15}{2} = -8$ $-\dfrac{1}{4}y = -\dfrac{1}{2}$ $y = 2$ Substitute $2$ for $y$ in the top equation. $-5x+ 2 = -8$ $-5x+2 = -8$ $-5x = -10$ $x = 2$ The solution is $\enspace x = 2, \enspace y = 2$.